A Piece of Pi For Dessert During Covid Shutdown

[Garland Best]

Here are few links:

https://www.khanacademy.org/math/mat...oof-infinities

https://www.askamathematician.com/20...20at%20random.

https://www.angio.net/pi/piquery

[Garland Best]

Using Cantor's method, I will generate an infinite string of digits that cannot exist in Pi.

1) Generate a list of all possible sequences of digits in Pi as follows.
a) item 1 in our list starts from the first spot after the decimal place: 141592653589793238462643383...
b) item 2 in our list starts from the second spot after the decimal place: 415926535897932384626433832...
c) item 3 in our list starts from the third spot after the decimal place: 159265358979323846264338327...
d) item 4 in our list starts from the fourth spot after the decimal place: 592653589793238462643383279...
e) item 5 in our list starts from the fifth spot after the decimal place: 926535897932384626433832795...
and so on.

2) Generate our new number by taking the nth digit in each number in our list (so 1 from item 1, 1 from item 2, 9 from item 3, 6 from item 4, 3 from item 1), and increasing the value by one. The pattern below will make this clear.
141592653589793238462643383...
415926535897932384626433832...
159265358979323846264338327...
592653589793238462643383279...
926535897932384626433832795...
.....
22074....

This new number:
a) Does not match the first number, because the first digit is different
b) Does not match the second number, because the second digit is different
c) Does not match the third number, because the third digit is different
.....
n) Does not match the nth number because the nth digit is different.

And onwards to infinity. Therefore this infinite string of digits that I created CANNOT exist in Pi. And I can create similar numbers using any number of different ways, as long as I ensure that given that the nth digit is different than the matching digit in the list.

Given that I can create an infinite number of infinite strings of digits that I can prove do not exist in Pi, it stands to reason that your pattern of infinite 7's does not have to exist in Pi.

[Garland Best]

So my question to Garland Best has been answered: pi must contain 777,777,777,777 consecutive 7's and it must contain it an infinite number of times.

Which leads me to wonder, if it contains it an infinite number of times, doesn't that imply that somewhere it must contain them BACK TO BACK TO BACK etc. an infinite number of times... ???

You are making two errors, the first open to debate, the second more obvious.

1) Nature or randomness: Since the digits of Pi are a sequence of digits generated by a formula, it is not an actual random sequence. Mathematicans still debate if the sequence truly meets the requirement of randomness, although so far it has passed all tests. So for now I will give you the point that it is possible that 777,777,777,777 consecutive 7's can occur within the digits of Pi. However

2) Again, do not confuse an arbitrarily long finite sequence with an infinite sequence. They are not the same. No matter how big a string of 7's you can come up with, I can come up with a bigger one and an infinite string of 7's will still be infinitely times larger.

[Pargat Perrer]

From my previous post, a math nerd (Master's degree) told me:

This logical inference isn't true: "Given that Pi goes on for infinity with no repeating pattern, we can say that any finite set of digits we can imagine should be contained somewhere within Pi"

After I read last night about Infinite Monkey theory, Aris, I believe your math nerd friend is in error. The Infinite Monkey Theory is summed up as follows:

"A monkey hitting typewriter keys at random for an infinite time will almost surely type any given text, such as the complete works of William Shakespeare. In fact, the monkey will almost surely type every possible finite text an infinite number of times."

The term "almost surely" is a mathematical term which means "probability = 1", which itself means 100% probability.

Since pi is generating digits randomly, it substitutes for the monkey, and the typewriter only contains digits 0 to 9 inclusive.

Therefore the statement of mine you quoted above is in fact true. And in the writeup for Infinite Monkey theory on Wikipedia, this is confirmed later down in the text:

"This is an extension of the principle that a finite string of random text has a lower and lower probability of being a particular string the longer it is (though all specific strings are equally unlikely). This probability approaches 0 as the string approaches infinity. Thus, the probability of the monkey typing an endlessly long string, such as all of the digits of pi in order, on a 90-key keyboard is (1/90)^∞

which equals (1/∞) which is essentially 0. At the same time, the probability that the sequence

contains a particular subsequence (such as the word MONKEY, or the 12th through 999th digits of pi, or a version of the King James Bible) increases as the total string increases. This probability approaches 1 as the total string approaches infinity, and thus the original theorem is correct. "

So my question to Garland Best has been answered: pi must contain 777,777,777,777 consecutive 7's and it must contain it an infinite number of times.

Which leads me to wonder, if it contains it an infinite number of times, doesn't that imply that somewhere it must contain them BACK TO BACK TO BACK etc. an infinite number of times... ???

[Bob Gillanders]

Well, I have a degree in Mathematics and enjoy a keen interest in astrophysics, so naturally I began reading this thread.
But then my pattern recognizing skills (thanks you chess) began to kick in as the post lengths began increasing to infinity with no repeating pattern, so I stopped reading.
Sanity preserved, for now.

[Aris Marghetis]

One thing I can ask is this: if instead I conjectured that somewhere within pi there must be a string of 777,777,777,777 consecutive 7's, would that be true?

From my previous post, a math nerd (Master's degree) told me:

This logical inference isn't true: "Given that Pi goes on for infinity with no repeating pattern, we can say that any finite set of digits we can imagine should be contained somewhere within Pi"

[Pargat Perrer]

Nope. Real thing. The set of natural numbers (1,2,3,4,5, and so on) is a countable infinite set, because each element can be placed in order, one after another, forming a list that can be "counted" to infinity. So are the set of whole numbers, integers, and even rational numbers (fractions). All of these numbers can be arranged in a list. So can the digits of pi. 3.1415926... is a list of the digits of pi.

On the other hand the set of REAL numbers, which include not just all rational numbers but also all irrational numbers as well (Pi, e, square root of 2, and so on) is an uncountable infinite set. Cantor proved that no matter how you try to arrange a list of real numbers, even just those between 0 and 1, there will always be numbers that are definitely real numbers, but are not in the list. There are lot of examples of this that you can find.

PS: You can use cantor's proof to come up with a infinite sequence of numbers that CANNOT possibly exist in PI. I will leave it to you to review Cantor's proof to come up with an example.

Thanks for that explanation Garland. I will do research on Cantor's proof. I hope it's something I can understand, not being a mathematician... :o

I think I do understand now what you mean by "countable". Again, I related it to computer programming, where if we create a class of some object type, and we want to have a list of objects of this class where the objects are in some order, we have to give the class a < (less than) function and an == function. So by "countable" we are really saying "orderable", is that correct?

One thing I can ask is this: if instead I conjectured that somewhere within pi there must be a string of 777,777,777,777 consecutive 7's, would that be true?

[Garland Best]

I had never heard the term "countable infinity" before and it seems like some artificial idea to try and imagine infinity. I would like to see some sort of understandable proof of what you say, that pi cannot contain an infinite number of infinities. Otherwise it just seems like an assertion meant to avoid thinking about it.

Nope. Real thing. The set of natural numbers (1,2,3,4,5, and so on) is a countable infinite set, because each element can be placed in order, one after another, forming a list that can be "counted" to infinity. So are the set of whole numbers, integers, and even rational numbers (fractions). All of these numbers can be arranged in a list. So can the digits of pi. 3.1415926... is a list of the digits of pi.

On the other hand the set of REAL numbers, which include not just all rational numbers but also all irrational numbers as well (Pi, e, square root of 2, and so on) is an uncountable infinite set. Cantor proved that no matter how you try to arrange a list of real numbers, even just those between 0 and 1, there will always be numbers that are definitely real numbers, but are not in the list. There are lot of examples of this that you can find.

PS: You can use cantor's proof to come up with a infinite sequence of numbers that CANNOT possibly exist in PI. I will leave it to you to review Cantor's proof to come up with an example.

[Pargat Perrer]

Here is your issue. This statement cannot be true. Only finite length sequences can occur. The sequence of digits in pi constitute a "countable infinity" and cannot contain an infinite number of infinities inside it.

I had never heard the term "countable infinity" before and it seems like some artificial idea to try and imagine infinity. I would like to see some sort of understandable proof of what you say, that pi cannot contain an infinite number of infinities. Otherwise it just seems like an assertion meant to avoid thinking about it.

So let's describe two separate things: a FINITE random sequence and its probability of containing a finite sequence of repeating single digits (approaches 0 probability as length of the repeating digits grows), and an INFINITE random sequence of digits and its probability of containing that same sequence of repeating digits (but this time, we increase the size of the random sequence towards infinity, leaving the repeating sequence length constant and finite, and as we do so, the probability approaches 1).

What the above doesn't address is the concept of increasing the length of the random sequence towards infinity (which increases probability of finding any specific finite subsequence within it) WHILE AT THE SAME TIME increasing the length of the string of repeating digits also towards infinity (which by itself, increasing this length decreases the probability of its occurrence within the random sequence towards 0). The first probability is increasing towards 1, while the other is decreasing towards 0.

If the random sequence truly never ends, then it must contain every conceivable finite subsequence within it. But if the length of repeating digits also goes on for infinity, it's probability of occurring within the random sequence approaches BUT NEVER REACHES 0. Since it never reaches 0, it must occur if the random sequence also never ends. The key concept here is PROBABILITY NEVER REACHES 0.

In computer programming there is the concept of a recursive function, that is, a function that calls itself. Such a function is normally given a termination condition. If it didnt' have that, the function would continue to call itself until the host computer's function stack filled all available memory and you have stack overflow which halts the thread. If you had infinite computer resources, the function could continue to call itself to infinity. But if we're imagining infinite computer resources, then on another thread, the argument to the recursive function could be calculated such that it isn't repeating itself ever, meaning each call of the function has a unique argument. If that is true, then there must be a call to the function that receives as an argument an infinite string of repeating digits.

This to me is part of imagining what infinity really is. All things are both possible AND required.

[Aris Marghetis]

So I was uncomfortable with some of the initial statements, so I asked a Master's in math about this, and this is his reply:



This logical inference isn't true: "Given that Pi goes on for infinity with no repeating pattern, we can say that any finite set of digits we can imagine should be contained somewhere within Pi"

As a consequence, it's not necessarily true that "ALL POSSIBLE STRINGS OF CONSECUTIVE DIGITS SHOULD BE PRESENT, INCLUDING INFINITE SEQUENCES OF ANY ONE PARTICULAR DIGIT."

The thing about non-terminating, non-repeating decimal expansions of irrational numbers is that they can be non-repeating in pretty boring, structured ways. They can still have patterns — the claim is that they are not generated by a repeating finite sequence

Consider the number, "0.10110111011110111110111110..." — the number of 1s increases incrementally — this is the same kind of number as Pi (non-terminating, non-repeating decimal expansion), since you can't generate it by just repeating a finite sequence. It also contains lots of instances of "11" — in fact, that particular sequence shows up an infinite number of times. You can't have an infinite sequence of digits that doesn't have repeated runs of digits, because there are only a finite number of digits (0-9) and thus only a finite number of ways of combining, say, two digits together.

So, the first inference by the poster, quoted above, just isn't true in general, and I think reflects a misunderstanding of what kind of 'thing' Pi is supposed to be. Non-repeating decimal expansions can consist of a very specific (but still infinite!) subset of strings, and they will necessarily contain some repeated sequences. What they won't be is generated by an infinitely repeated finite sequence — unlike, for instance, 0.1234123412341234

Having said that — it's not necessarily true that non-terminating, non-repeating numbers don't contain every possible finite sequence. They actually might! For instance, if an infinite sequence of digits were generated by drawing digits randomly (each digit equally likely), then in principle that sequence will contain every possible sequence of any arbitrary length. You might just need to "wait a long time" :p

In the case of Pi, it's actually an open area of research whether it contains every arbitrary finite sequence! A mathematician would gain fame and... well, not fortune, but fame if they were to prove it to be true or false. I think most mathematicians suspect it's actually true that the decimal expansion of Pi contains every finite sequence, but they're not sure how to prove it.

[Garland Best]

"This is simply because within an infinite sequence of digits, ALL POSSIBLE STRINGS OF CONSECUTIVE DIGITS SHOULD BE PRESENT, INCLUDING INFINITE SEQUENCES OF ANY ONE PARTICULAR DIGIT."

Here is your issue. This statement cannot be true. Only finite length sequences can occur. The sequence of digits in pi constitute a "countable infinity" and cannot contain an infinite number of infinities inside it.

[Dilip Panjwani]

Almost everything [including Pi :-)] began after the universal 'big bang' and will end before the universal 'black hole'; so nothing (except the cycle of 'big bangs' and 'black holes') is infinite. Trying to make sense of infinity as it applies to anything (except this cycle) can only make you go mad :-), as the fabled occurrence with Cantor...

[Stephen Humphreys]

There is a legend in the math world that this sort of paradox is what drove Georg Cantor mad. First Covid-19 and now this...

https://www.quora.com/Is-it-true-tha...magine?share=1

[Pargat Perrer]

As we continue to experience "covid fever" (an offshoot of cabin fever) I thought I'd like to introduce a mathematical conundrum regarding the number Pi for all of us to waste our time on.

As you most likely know, Pi is a fractional number with no end. It goes on for infinity with theoretically no repeating pattern whatsoever. This conundrum could apply to any such number, but let's stick to Pi.

I've read that Pi has been generated up to and beyond 10 trillion digits (after the decimal point, starting with 3.14159....).

Given that Pi goes on for infinity with no repeating pattern, we can say that any finite set of digits we can imagine should be contained somewhere within Pi. For example, somewhere within Pi must exist the set of digits "7777777" (seven consecutive 7's). Similarly, somewhere within Pi (might be somewhere else entirely) there should be eight consecutive 7's. And somewhere else nine consecutive 7's. And somewhere else 900 consecutive 7's. 9 thousand consecutive 7's. 9 million consecutive 7's.... you get the idea.

In fact, somewhere within Pi there should be infinite consecutive 7's (or any other digit from 0 to 9 you want to propose). This is simply because within an infinite sequence of digits, ALL POSSIBLE STRINGS OF CONSECUTIVE DIGITS SHOULD BE PRESENT, INCLUDING INFINITE SEQUENCES OF ANY ONE PARTICULAR DIGIT.

BUT.... if indeed there were to be found infinite consecutive 7's or any other digit, isn't that a repeating pattern? How can we have an infinitely long repeating pattern within an infinitely long non-repeating pattern?

But here's an even bigger conundrum.... if there must be an infinite string of consecutive 7's.... there must also be an infinite string of consecutive 8's. How can two consecutive infinite strings appear within a single infinite non-repeating string of digits? If the 7's appear first, and they go on for infinity, where do the 8's go?

And here's an even bigger question.... if there must be infinite consecutive 7's and infinite consecutive 8's within Pi, then there must also be infinite consecutive 0's!

But infinite consecutive 0's means we've reached the end of the number! It's not infinite after all!